Example 1.3.34 on page 74 of Ilya Molchanov's Theory of Random Sets makes the claim that if $(\xi(\omega,t);\omega\in\varOmega,t\geq0)$ is a (jointly measurable) stochastic process, then its path $\varXi = \{\xi(\cdot,t)\colon t\geq0\}$ is graph measurable.

By graph measurable, the author means that the set $\{(\omega,r)\in\varOmega\times\mathbb{R}\colon r\in\varXi(\omega)\}$ is in the product $\sigma$-algebra $\mathcal{A}\otimes\mathcal{B}(\mathbb{R})$, where $\mathcal{A}$ is a $\sigma$-algebra on the sample space $\varOmega$ and $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$.

This is how I tried proceeding:

$$\{(\omega,r)\in\varOmega\times\mathbb{R}\colon r\in\varXi(\omega)\} = \{(\omega,r)\in\varOmega\times\mathbb{R}\colon \exists t\geq0~\text{such that}~\xi(\omega,t) = r\} = \bigcup_{t\geq0}\{(\omega,r)\in\varOmega\times\mathbb{R}\colon \xi(\omega,t) = r\}.$$

Now, for each fixed $t$, the set $\{(\omega,r)\in\varOmega\times\mathbb{R}\colon \xi(\omega,t) = r\}$ is easily shown to be in $\mathcal{A}\otimes\mathcal{B}(\mathbb{R})$. But, without any further assumptions, I don't really see why that uncountable union over the $t$-s would also be in the product $\sigma$-algebra.

And, in addition, if this claim is not true in general, what assumptions (path continuity or, may be, separability??) can be imposed on the process $\xi$ to make it happen?


I can tell you it is not true in general.


Let $\mathcal{A}=\{\emptyset,\Omega\}$ and define the measurable function $f:[0,1]\to[0,1]$ be as in this answer, whose image $E=\{f(t):t\in[0,1]\}$ is not measurable. Define, for all $\omega\in\Omega$, $\xi(\omega,t)=f(\max\{t,1\})$. The measurability of $f$ implies that of $\xi$. However, $\Xi(\omega)=\{f(t):t\in[0,1]\}=E$ and $\{(\omega,r)\in\Omega\times\mathbb{R} :r\in\Xi(\omega)\}=\Omega\times E$, which is not measurable.

Sufficient conditions

Let $\mathbb{P}$ be a probability measure. If $\xi(\omega,\cdot)$ is separable on $\Omega\setminus\Omega_0$ where $\Omega_0$ is $\mathbb{P}$-null, and if $\mathcal{A}$ contains all subsets of $\Omega_0$, then a modification of the closure $\overline{\Xi}$ is graph-measurable.

Indeed, let $\xi'$ be the indistinguishable from $\xi$, defined so that they agree on $\Omega$ and $\xi'$ is constant and equal to $0$ on $\Omega_0$ (note that $\xi'$ remains jointly measurable, by assumption). Let $T\subset[0,\infty)$ be the countable set that separates $\xi$. Define, for $n\geq 1$ and $k\in\mathbb{Z}$, the interval $I_{k,n}=\big[\frac{k-1}{2^n},\frac{k+1}{2^n}\big]$. Note that

$$A:=\{(\omega,r)\in\Omega\times\mathbb{R}: r\in\overline{\Xi'(\omega)}\}= \bigcap_{n\in\mathbb{N}}\bigcup_{k\in\mathbb{Z}}\bigcup_{t\in T} \big\{\omega\in\Omega:\big|\xi'(\omega,t)-k2^{-n}\big|\leq 2^{-n}\big\}\times I_{k,n}.$$

(Indeed, $r$ is arbitrarily close to some value of $\xi'(\omega,\cdot)$ if and only if it is arbitrarily close to (at most $2^{-n}$ for any $n$) some diadic rational (here $k2^{-n}$) which is itself just as close to some value of $\{\xi'(\omega,t):t\in T\}$.) The measurability of $A$ is clear since (I) these are countable unions and intersection and (II) $\big\{\omega\in\Omega: \big|\xi'(\omega,t)-k2^{-n}\big|\leq 2^{-n}\big\}$ is measurable (which follows easily from the joint measurability of $\xi'$) and hence so are the rectangles.

I hope this helps.

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